How to Complete the Square for a Diffusion Equation?

[trap101]

Ok this qusestion has to do with completing the square for a diffusion equation.

Initial Cond: u(x,0) = e^-x

Now they say plug this into the general formula:

u(x,t) = 1/(4$\pi$kt)^1/2 ∫ e^{-(x-y)1/2/4kt}e^-y dy where k is a constant

now the first step they say is completing the square of:

-( x²-2xy+y²)+4kty/4kt

with respect to the y variable, and they get:

- [(y+2kt-x)²]/4kt + kt - x

Now I could not get this, I also tried expanding out the final result and reverse engineer the result but in doing so I got stuck with an extra term:

y²+ 2y(2kt-x) + x² + (2kt-x)² - (2kt-x)²

this step is when I perform the process of completing the square before trying to factorize everything and it is here that I am having trouble. Please help if you can I have the midterm in a couple hrs.

Thanks

 

[LCKurtz]

Ok this qusestion has to do with completing the square for a diffusion equation.

Initial Cond: u(x,0) = e^-x

Now they say plug this into the general formula:

u(x,t) = 1/(4$\pi$kt)^1/2 ∫ e^{-(x-y)1/2/4kt}e^-y dy where k is a constant

now the first step they say is completing the square of:

-( x²-2xy+y²)+4kty/4kt

with respect to the y variable, and they get:

- [(y+2kt-x)²]/4kt + kt - x

The first thing is to state the problem correctly:
$$\frac{-(x^2-2xy+y^2)+4kty}{4kt}$$which is not what you have written. Rearrange the numerator to$$
-(y^2+(2x-4kt)y)-x^2$$Now complete the square, not forgetting the - out in front$$
-(y^2+(2x-4kt)y+(x-2kt)^2)+(x-2kt)^2-x^2$$ $$
=-(y+(x-2kt))^2+x^2-4kxt+4t^2-x^2$$Now put the denominator back in and simplify it.

 

[tiny-tim]

hi trap101!

-( x²-2xy+y²)+4kty/4kt

with respect to the y variable, and they get:

- [(y+2kt-x)²]/4kt + kt - x

[(y+2kt-x)²]/4kt

= -[y² + 4k²t² + x² - 2xy + 4kty - 4ktx]/kt + kt - x